October/November 2025 Paper 41 Worked Answers (IGCSE Maths 0580 Extended)

46 questions · 100 marks · 120 minutes

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1. Question 1

   1 marksProperties of quadrilaterals
   Step 1: A quadrilateral with rotational symmetry of order 2 could be a rectangle, rhombus, or parallelogram. Step 2: Having exactly two lines of symmetry that are its diagonals uniquely identifies a rhombus. Step 3: A rectangle has two lines of symmetry along its midpoints, not its diagonals. A parallelogram has no lines of symmetry. A square has 4 lines of symmetry.
   Method:

   Identify which quadrilateral has exactly two lines of symmetry that are its diagonals and rotational symmetry of order 2. This is the rhombus.
   Examiner tips

   • Know the symmetry properties of all standard quadrilaterals
   • Distinguish between diagonals and perpendicular bisectors as lines of symmetry

2. Question 2

   2 marksSolving linear equations
   Step 1: Rearrange: $11 - 4 = 2x$, so $7 = 2x$. Step 2: Divide both sides by 2: $x = \frac{7}{2} = 3.5$.
   Method:

   Subtract 4 from both sides to get 7 = 2x, then divide by 2 to get x = 3.5.
   Examiner tips

   • Show each step of rearrangement clearly
   • Check your answer by substituting back into the original equation

3. Question 3

   2 marksAngles in parallel lines
   Step 1: The $55°$ angle and its co-interior angle on the same side of the transversal are supplementary (co-interior angles add up to $180°$). Step 2: $x = 180 - 55 = 125°$.
   Method:

   Recognise the angles are co-interior angles between parallel lines, so x + 55 = 180, giving x = 125.
   Examiner tips

   • Identify whether the angles are alternate, corresponding, or co-interior before calculating
   • Co-interior angles sum to 180 degrees

4. Question 4

   1 marksTime calculations
   Step 1: Add the hours: $22\text{:}16 + 5\text{ hours} = 03\text{:}16$ (next day, since $22 + 5 = 27$, and $27 - 24 = 3$). Step 2: Add the minutes: $03\text{:}16 + 52\text{ minutes} = 04\text{:}08$ (since $16 + 52 = 68$ minutes $= 1$ hour $8$ minutes).
   Method:

   Add 5 hours 52 minutes to 22:16. The hours give 03:16 (next day), then adding 52 minutes gives 04:08.
   Examiner tips

   • Be careful with times that cross midnight
   • Convert minutes greater than 60 into hours and minutes

5. Question 5a

   2 marksConstructions
   Step 1: Check the triangle inequality: the sum of any two sides must be greater than the third. Step 2: $8 + 6 = 14 > 11$ ✓, $11 + 6 = 17 > 8$ ✓, $11 + 8 = 19 > 6$ ✓. Step 3: The triangle can be constructed.
   Method:

   Draw arcs of radius 7 cm from A and 5 cm from B. Their intersection is C. Join AC and BC to complete the triangle.
   Examiner tips

   • Always leave your construction arcs visible
   • Check that the lengths of all three sides match the given measurements

6. Question 5b

   1 marksMeasuring angles
   Step 1: Angles in a triangle sum to $180°$. Step 2: Angle ACB = $180° - 52° - 45° = 83°$.
   Method:

   Use a protractor to measure angle ACB at vertex C. The answer should be between 80 and 85 degrees.
   Examiner tips

   • Make sure the protractor centre is exactly on the vertex
   • Read the correct scale on the protractor

7. Question 5c_i

   1 marksScale drawings
   Step 1: $AB = 8$ cm on the drawing. Step 2: Actual distance $= 8 \times 10\,000 = 80\,000$ cm $= 800$ m $= 0.8$ km.
   Method:

   Multiply 8 cm by 10000 to get 80000 cm, then convert to km: 80000 / 100000 = 0.8 km.
   Examiner tips

   • Remember: 1 km = 100,000 cm
   • Apply the scale factor first, then convert units

8. Question 5c_ii

   1 marksBearings
   Step 1: If $B$ is due east of $A$, then $A$ is due west of $B$. Step 2: The bearing of due west is $270°$.
   Method:

   B is due east of A, so A is due west of B. The bearing of due west is 270 degrees.
   Examiner tips

   • Read the question carefully: bearing OF A FROM B means stand at B and look towards A
   • Due west is 270 degrees

9. Question 6

   1 marksExpanding brackets
   Step 1: Multiply $g$ by each term inside the bracket: $g \times 3 = 3g$ $g \times (-2g) = -2g^{2}$ Step 2: Result: $3g - 2g^{2}$.
   Method:

   Distribute g across both terms: g × 3 = 3g and g × (-2g) = -2g², giving 3g - 2g².
   Examiner tips

   • Multiply every term inside the bracket by the term outside
   • Remember that g × g = g²

10. Question 7a

    2 marksInequalities on a number line
    Step 1: An open circle at $-3$ means $x > -3$ (not equal to $-3$). Step 2: A closed circle at $2$ means $x \leq 2$ (can equal $2$). Step 3: Combined: $-3 < x \leq 2$.
    Method:

    Read the number line: open circle at -3 gives strict inequality, closed circle at 2 gives inclusive inequality, so -3 < x ≤ 2.
    Examiner tips

    • Open circle = strict inequality (< or >)
    • Closed circle = inclusive inequality (≤ or ≥)

11. Question 7b

    2 marksInteger solutions of inequalities
    Step 1: Divide all parts by 2: $-2 < x \leq 4$. Step 2: List integers satisfying this: $x = -1, 0, 1, 2, 3, 4$. Note: $x = -2$ is not included because the inequality is strict ($<$, not $\leq$). $x = 4$ is included because the inequality is inclusive ($\leq$).
    Method:

    Divide by 2 to get -2 < x ≤ 4, then list all integers from -1 to 4 inclusive.
    Examiner tips

    • Simplify the inequality first by dividing by 2
    • Be careful whether each endpoint is included or excluded

12. Question 8

    1 marksCalculator computation
    Step 1: $3.5^{2} = 12.25$. Step 2: $2.2^{3} = 10.648$. Step 3: $12.25 - 10.648 = 1.602$. Step 4: $1.602^{0.25} = 1.125\ldots \approx 1.13$.
    Method:

    Compute 3.5² = 12.25, then 2.2³ = 10.648, subtract to get 1.602, then take the fourth root to get 1.125... ≈ 1.13.
    Examiner tips

    • Use your calculator carefully, paying attention to the order of operations
    • A fractional power of 1/4 means the fourth root

13. Question 9

    2 marksUnit conversion (speed)
    Step 1: Convert m/s to km/h by multiplying by $\frac{3600}{1000} = 3.6$. Step 2: $9.5 \times 3.6 = 34.2$ km/h.
    Method:

    Multiply 9.5 by 3.6 to convert from m/s to km/h: 9.5 × 3.6 = 34.2 km/h.
    Examiner tips

    • Remember: m/s to km/h multiply by 3.6
    • You can also multiply by 3600 (seconds in an hour) then divide by 1000 (metres in a km)

14. Question 10

    2 marksColumn vectors
    Step 1: $B = A + \vec{AB}$. Step 2: $B = (2 + 4,\; 1 + 2) = (6, 3)$.
    Method:

    Add the vector (4, 2) to point A(2, 1): B = (2+4, 1+2) = (6, 3).
    Examiner tips

    • To find B, add the vector components to the coordinates of A
    • Check: the vector from A to B should match the given vector

15. Question 11a

    2 marksPercentage decrease
    Step 1: Calculate the discount: $15\%$ of $\$60 = 0.15 \times 60 = \$9$. Step 2: Sale price $= \$60 - \$9 = \$51$. Alternatively: $60 \times 0.85 = \$51$.
    Method:

    Sale price = original price \times (1 - discount\%). Set up the equation and solve for the original price.
    Examiner tips

    • To decrease by 15%, multiply by 0.85 or find 15% and subtract
    • Make sure you give the sale price, not the discount

16. Question 11b

    2 marksReverse percentages
    Step 1: The sale price is 85% of the original price. Step 2: Original price $= \frac{58.14}{0.85} = \$68.40$.
    Method:

    Sale price is 85% of original. Original = 58.14 ÷ 0.85 = $68.40.$
    Examiner tips

    • In reverse percentage problems, divide by the multiplier (0.85), do not add 15% to the sale price
    • The sale price is 85% of the original, not 100%

17. Question 12a

    1 marksVenn diagrams
    Step 1: Students who do not like rugby are those in $F$ only and those outside both sets. Step 2: $33 + 4 = 37$.
    Method:

    Add the regions outside R: F only (33) + outside both (4) = 37.
    Examiner tips

    • Not liking rugby means outside the entire R circle
    • Include those outside both sets as well as those only in F

18. Question 12b

    1 marksSet notation
    Step 1: 'Like rugby' means in set $R$. Step 2: 'But not football' means not in set $F$, i.e., in $F'$. Step 3: The intersection gives $F' \cap R$.
    Method:

    Rugby but not football means in R and in F', so the set notation is F' ∩ R.
    Examiner tips

    • Use ∩ for 'and' and ' (prime) for 'not'
    • Read the description carefully: rugby BUT NOT football

19. Question 13

    3 marksAngle of elevation
    Step 1: The vertical difference between $T$ and $E$ is $12.6 - 1.5 = 11.1$ m. Step 2: The horizontal distance is $33$ m. Step 3: $\tan(\theta) = \frac{11.1}{33}$. Step 4: $\theta = \tan^{-1}\left(\frac{11.1}{33}\right) = 18.6°$.
    Method:

    Height difference = 12.6 - 1.5 = 11.1 m. Horizontal distance = 33 m. Angle = tan⁻¹(11.1/33) = 18.6°.
    Examiner tips

    • Subtract the heights of the two poles to find the vertical difference
    • The angle of elevation is measured from the horizontal at E looking up to T

20. Question 14a

    1 marksVolume of a sphere
    Step 1: The sphere touches all faces of the cube, so the diameter equals the side length: $d = 8$ cm, thus $r = 4$ cm. Step 2: $V = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi(4)^{3} = \frac{4}{3}\pi \times 64 = \frac{256}{3}\pi$ cm$^{3}$.
    Method:

    Radius = 8/2 = 4 cm. Volume = 4/3 × π × 4³ = 4/3 × π × 64 = 256π/3 cm³.
    Examiner tips

    • In 'show that' questions, show every step of working
    • State that the radius is half the side length of the cube

21. Question 14b

    3 marksPercentage of volume
    Step 1: Volume of cube $= 8^{3} = 512$ cm$^{3}$. Step 2: Volume of sphere $= \frac{256}{3}\pi \approx 268.08$ cm$^{3}$. Step 3: Volume not occupied $= 512 - 268.08 = 243.92$ cm$^{3}$. Step 4: Percentage $= \frac{243.92}{512} \times 100 \approx 47.6\%$.
    Method:

    Cube volume = 512. Unoccupied = 512 - 256π/3. Percentage = (512 - 256π/3)/512 × 100 ≈ 47.6%.
    Examiner tips

    • Read carefully whether the question asks for occupied or not occupied
    • Show the volume of the cube and the subtraction clearly

22. Question 14c

    2 marksDensity calculation
    Step 1: Mass $=$ density $\times$ volume $= 7.86 \times \frac{256}{3}\pi \approx 7.86 \times 268.08 \approx 2107$ g. Step 2: Convert to kg: $2107 \div 1000 = 2.11$ kg.
    Method:

    Mass = 7.86 × 256π/3 = 2107 g = 2.11 kg.
    Examiner tips

    • Remember density = mass/volume, so mass = density × volume
    • Convert grams to kilograms by dividing by 1000

23. Question 14d

    4 marksSurface area of a cylinder
    Step 1: Volume of cylinder $=$ volume of sphere: $\pi(3.1)^{2}h = \frac{256}{3}\pi$. Step 2: $h = \frac{256}{3 \times 3.1^{2}} = \frac{256}{3 \times 9.61} = \frac{256}{28.83} = 8.880\ldots$ cm. Step 3: Total surface area $= 2\pi r^{2} + 2\pi r h = 2\pi(3.1)^{2} + 2\pi(3.1)(8.880)$. Step 4: $= 2\pi(9.61) + 2\pi(27.528) = 60.38 + 172.97 \approx 233$ cm$^{2}$.
    Method:

    Find h from πr²h = 256π/3, giving h = 256/(3 × 9.61) ≈ 8.88 cm. TSA = 2π(3.1)² + 2π(3.1)(8.88) ≈ 233 cm².
    Examiner tips

    • Equate the sphere volume to the cylinder volume to find the height
    • Total surface area of a cylinder = 2πr² + 2πrh

24. Question 15

    3 marksCircle theorems with tangents
    Step 1: Angle at centre $= 2 \times$ angle at circumference: $\angle AOB = 2 \times 67 = 134°$. Step 2: Tangent meets radius at $90°$: $\angle OAT = \angle OBT = 90°$. Step 3: In quadrilateral $OATB$: $\angle ATB = 360 - 90 - 90 - 134 = 46°$. Alternatively: $\angle ATB = 180 - 2 \times 67 = 46°$.
    Method:

    Angle AOB = 2 × 67 = 134°. In quadrilateral OATB, angle ATB = 360 - 90 - 90 - 134 = 46°.
    Examiner tips

    • The angle at the centre is twice the angle at the circumference
    • Tangent is perpendicular to the radius at the point of contact

25. Question 16

    4 marksEstimated mean from grouped data
    Step 1: Mid-values: $7.5$, $11$, $16$. Step 2: $\Sigma fx = 3 \times 7.5 + 24 \times 11 + 23 \times 16 = 22.5 + 264 + 368 = 654.5$. Step 3: Mean $= \frac{654.5}{50} = 13.09$ cm.
    Method:

    Mid-values: 7.5, 11, 16. Σfx = 3(7.5) + 24(11) + 23(16) = 654.5. Mean = 654.5/50 = 13.09 cm.
    Examiner tips

    • Always use the mid-value of each class for estimating the mean
    • Show your Σfx calculation clearly

26. Question 17

    3 marksEquation of a straight line
    Step 1: Gradient $m = \frac{4 - 0}{0 - 2} = \frac{4}{-2} = -2$. Step 2: The $y$-intercept is $c = 4$ (since the line passes through $(0, 4)$). Step 3: $y = -2x + 4$.
    Method:

    Gradient = (4-0)/(0-2) = -2. y-intercept = 4. Equation: y = -2x + 4.
    Examiner tips

    • The point (0, 4) gives you the y-intercept directly
    • Be careful with the sign of the gradient

27. Question 18a

    2 marksSpeed-time graphs: distance
    Step 1: Distance = area under the speed-time graph. Step 2: Triangle (0 to 2 s): $\frac{1}{2} \times 2 \times 8 = 8$ m. Step 3: Rectangle (2 to 10 s): $8 \times 8 = 64$ m. Step 4: Total $= 8 + 64 = 72$ m.
    Method:

    Area = ½(2)(8) + 8(8) = 8 + 64 = 72 m.
    Examiner tips

    • Distance = area under a speed-time graph
    • Split complex shapes into triangles and rectangles

28. Question 18b

    2 marksSpeed-time graphs: total time
    Step 1: Distance remaining $= 100 - 72 = 28$ m. Step 2: Time for remaining distance $= \frac{28}{8} = 3.5$ s. Step 3: Total time $= 10 + 3.5 = 13.5$ s.
    Method:

    Remaining = 100 - 72 = 28 m. Extra time = 28/8 = 3.5 s. Total = 10 + 3.5 = 13.5 s.
    Examiner tips

    • Use the distance from part (a) to find the remaining distance
    • Remember to add the initial 10 seconds to get the total time

29. Question 19

    3 marksExponential decay
    Step 1: After $n$ days: mass $= 20 \times 0.9^{n}$. Step 2: Solve $20 \times 0.9^{n} < 1$, so $0.9^{n} < 0.05$. Step 3: $n > \frac{\ln(0.05)}{\ln(0.9)} = \frac{-2.996}{-0.1054} = 28.43\ldots$ Step 4: So $n = 29$ whole days.
    Method:

    20 × 0.9^n < 1, so 0.9^n < 0.05. Using logarithms: n > 28.43, so n = 29 whole days.
    Examiner tips

    • 10% decay means the multiplier is 0.9, not 0.1
    • The answer must be a whole number of days, so round up

30. Question 20a

    1 marksGeometric sequences
    Step 1: Each term is half the previous term (common ratio $= \frac{1}{2}$). Step 2: Next term $= 3 \times \frac{1}{2} = 1.5$.
    Method:

    Common ratio = ½. Next term = 3 × ½ = 1.5.
    Examiner tips

    • Check whether the sequence is arithmetic (common difference) or geometric (common ratio)
    • Each term is halved, so the common ratio is ½

31. Question 20b

    2 marksnth term of geometric sequence
    Step 1: The $n$th term of a geometric sequence is $ar^{n-1}$ where $a = 48$ and $r = \frac{1}{2}$. Step 2: $T_{n} = 48 \times \left(\frac{1}{2}\right)^{n-1} = 48 \times \frac{1}{2^{n-1}} = \frac{48}{2^{n-1}} = \frac{96}{2^{n}} = 96 \times \left(\frac{1}{2}\right)^{n}$.
    Method:

    a = 48, r = ½. nth term = 48 × (½)^(n-1) = 96 × (½)^n.
    Examiner tips

    • Check your formula gives the correct first term when n = 1
    • The standard form is ar^(n-1) but equivalent forms are accepted

32. Question 21

    2 marksArea of triangle using sine
    Step 1: Area $= \frac{1}{2}ab\sin C = \frac{1}{2} \times 8 \times 9 \times \sin 50°$. Step 2: $= \frac{1}{2} \times 72 \times 0.766 = 27.6$ cm$^{2}$.
    Method:

    Area = ½ × 8 × 9 × \sin 50° = 36 × 0.766 = 27.6 cm².
    Examiner tips

    • Use the sine rule for area when you have two sides and the included angle
    • Do not forget the ½ in the formula

33. Question 22

    3 marksCompound interest
    Step 1: $200\left(1 + \frac{r}{100}\right)^{25} = 301.10$. Step 2: $\left(1 + \frac{r}{100}\right)^{25} = \frac{301.10}{200} = 1.5055$. Step 3: $1 + \frac{r}{100} = 1.5055^{1/25} = 1.01649\ldots$ Step 4: $\frac{r}{100} = 0.01649\ldots$, so $r = 1.65$.
    Method:

    200(1 + r/100)^25 = 301.10. (1 + r/100)^25 = 1.5055. Take 25th root: 1 + r/100 = 1.0165. r = 1.65.
    Examiner tips

    • Set up the compound interest formula first
    • Isolate the bracket, then take the appropriate root

34. Question 23

    2 marksProportion
    Step 1: $y \propto \frac{1}{\sqrt{x+1}}$, so $y = \frac{k}{\sqrt{x+1}}$. Step 2: When $x = 8$: $3 = \frac{k}{\sqrt{9}} = \frac{k}{3}$, so $k = 9$. Step 3: $y = \frac{9}{\sqrt{x+1}}$.
    Method:

    y = k/√(x+1). When x=8: 3 = k/√9 = k/3, so k = 9. Therefore y = 9/√(x+1).
    Examiner tips

    • Set up y = k/√(x+1) for inverse proportion
    • Substitute the given values to find k

35. Question 24a

    1 marksForming expressions
    Step 1: Time $=$ distance $\div$ speed $= \frac{10}{x}$ hours.
    Method:

    Time = distance / speed = 10/x hours.
    Examiner tips

    • Remember: time = distance / speed
    • Make sure the fraction is the right way up

36. Question 24b

    4 marksForming and simplifying equations
    Step 1: $\frac{10}{x} + \frac{5}{x+4} = \frac{7}{2}$. Step 2: Multiply through by $2x(x+4)$: $20(x+4) + 10x = 7x(x+4)$. Step 3: $20x + 80 + 10x = 7x^{2} + 28x$. Step 4: $30x + 80 = 7x^{2} + 28x$. Step 5: $0 = 7x^{2} - 2x - 80$.
    Method:

    10/x + 5/(x+4) = 7/2. Multiply by 2x(x+4): 20(x+4) + 10x = 7x(x+4). Expand: 20x+80+10x = 7x²+28x. Simplify: 7x²-2x-80 = 0.
    Examiner tips

    • Show every step clearly in a 'show that' question
    • Convert 3.5 to 7/2 before multiplying

37. Question 24c

    3 marksQuadratic formula
    Step 1: Using the quadratic formula: $x = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4(7)(-80)}}{2(7)}$. Step 2: $x = \frac{2 \pm \sqrt{4 + 2240}}{14} = \frac{2 \pm \sqrt{2244}}{14}$. Step 3: $\sqrt{2244} = 47.37\ldots$ Step 4: $x = \frac{2 + 47.37}{14} = 3.53$ or $x = \frac{2 - 47.37}{14} = -3.24$.
    Method:

    x = (2 ± √(4+2240))/14 = (2 ± √2244)/14. x = 3.53 or x = -3.24.
    Examiner tips

    • Show the full substitution into the quadratic formula
    • Give both roots even if one is negative

38. Question 24d

    3 marksProblem solving with quadratics
    Step 1: Time at 80 km/h $= \frac{120}{80} = 1.5$ hours. Step 2: Time at 60 km/h $= \frac{120}{60} = 2$ hours. Step 3: Difference $= 2 - 1.5 = 0.5$ hours $= 30$ minutes.
    Method:

    Walking time = 10/3.53 = 2.833 h. Running time = 5/7.53 = 0.664 h. Difference = 2.169 h = 2 h 10 min.
    Examiner tips

    • Use the positive root only since x represents speed
    • Convert decimal hours to minutes by multiplying the decimal part by 60

39. Question 25a

    1 marksHistograms: reading frequency
    Step 1: Frequency $=$ frequency density $\times$ class width $= 20 \times 2 = 40$.
    Method:

    Frequency = frequency density × class width = 20 × 2 = 40.
    Examiner tips

    • In a histogram, frequency = frequency density × class width
    • Check the class width carefully from the axis

40. Question 25b

    2 marksHistograms: completing
    Step 1: Frequency density $= \frac{\text{frequency}}{\text{class width}} = \frac{102}{5 - 2} = \frac{102}{3} = 34$. Step 2: Draw a bar from $m = 2$ to $m = 5$ with height $34$.
    Method:

    Frequency density = 102/3 = 34. Draw bar from 2 to 5 with height 34.
    Examiner tips

    • Frequency density = frequency / class width
    • The class width is 5 - 2 = 3, not 5

41. Question 26

    4 marksSimultaneous equations (one linear, one quadratic)
    Step 1: Set equal: $2x^{2} - 3x - 7 = 2x - 7$. Step 2: Simplify: $2x^{2} - 5x = 0$. Step 3: Factorise: $x(2x - 5) = 0$. Step 4: $x = 0$ or $x = 2.5$. Step 5: When $x = 0$: $y = 2(0) - 7 = -7$. When $x = 2.5$: $y = 2(2.5) - 7 = -2$.
    Method:

    2x² - 3x - 7 = 2x - 7 gives 2x² - 5x = 0, so x(2x-5) = 0. x = 0 (y = -7) or x = 2.5 (y = -2).
    Examiner tips

    • Always find both pairs of solutions for simultaneous equations
    • Use the simpler equation to find y-values

42. Question 27

    4 marks3D trigonometry
    Step 1: $G$ is directly above a point on the base. Since $ABFE$ is the base, $G$ is directly above a point that is $7$ cm from $F$ along $FE$ (since $DC = 7$) and $5$ cm above the base (since $AD = 5$). Step 2: The projection of $G$ onto the base is the point $G'$ where $G'$ is at position $(7, 0, 15)$ relative to $A$ at origin. So $AG' = \sqrt{7^{2} + 15^{2}} = \sqrt{49 + 225} = \sqrt{274}$. Step 3: The vertical height from $G'$ to $G$ is $5$ cm. Step 4: $\tan(\theta) = \frac{5}{\sqrt{274}}$. Step 5: $\theta = \tan^{-1}\left(\frac{5}{\sqrt{274}}\right) = \tan^{-1}(0.3021) = 16.8°$.
    Method:

    Base projection AG' = √(7² + 15²) = √274. Height = 5. Angle = tan⁻¹(5/√274) = 16.8°.
    Examiner tips

    • In 3D problems, identify the right-angled triangle by projecting onto the base
    • The angle with the base involves the vertical height and the base projection

43. Question 28a

    1 marksFunctions: solving f(x) = k
    Step 1: $7^{x-4} = 1$. Step 2: Since $7^{0} = 1$, we need $x - 4 = 0$. Step 3: $x = 4$.
    Method:

    7^(x-4) = 1 = 7^0, so x - 4 = 0, x = 4.
    Examiner tips

    • Remember that a^0 = 1 for any non-zero a
    • Set the exponent equal to 0

44. Question 28b

    2 marksInverse functions
    Step 1: If $f^{-1}(x) = 1$, then $x = f(1)$. Step 2: $f(1) = 7^{1-4} = 7^{-3} = \frac{1}{343}$.
    Method:

    f⁻¹(x) = 1 means x = f(1) = 7^(1-4) = 7^(-3) = 1/343.
    Examiner tips

    • f⁻¹(x) = 1 means x = f(1)
    • A negative index means the reciprocal

45. Question 29a

    3 marksCosine rule
    Step 1: Use the cosine rule: $\cos y = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC}$. Step 2: $\cos y = \frac{10^2 + 14^2 - 13^2}{2 \times 10 \times 14} = \frac{127}{280}$. Step 3: $y = \cos^{-1}(\frac{127}{280}) = 63.0°$.
    Method:

    cos y = (10² + 14² - 13²)/(2 × 10 × 14) = 127/280. y = cos⁻¹(0.4536) = 63.0°.
    Examiner tips

    • When you know three sides, use the cosine rule to find any angle
    • The side opposite the angle goes on the left of the equation

46. Question 29b

    5 marksSine rule and cosine rule combined
    Step 1: Use the cosine rule: $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(PQR)$. Step 2: $PR^2 = 11^2 + 8^2 - 2(11)(8)\cos 65° = 121 + 64 - 176 \times 0.4226 = 185 - 74.4 = 110.6$. Step 3: $PR = \sqrt{110.6} = 10.5$ cm.
    Method:

    In triangle ACD, use the sine rule: AD/sin(ACD) = AC/sin(ADC). Calculate the missing angle and apply the sine rule to find AD = 15.1 cm.
    Examiner tips

    • Break the problem into two triangles
    • Find AD first using the sine rule in triangle ACD, then use the cosine rule in triangle ABD